<p>A worker can lift a box a vertical distance of 1m or roll it up a two meter-long ramp to the same elevation. Either way he does the same amount of work. If he uses the ramp, the applied force required is how much of the force that if he doesn't use the ramp? </p>
<p>A person uses 4.2J of work to push a 0.6kg box on a 30 degree slope, what's the distance that the box travels on the slope? </p>
<p>A Santa Claus looks at a spherical ball, what can you conclude about the image that he sees?</p>
<p>I’ll give it a go, but you’ll want to check the answers in class since I may be doing this wrong:</p>
<p>The work the person must perform in the first case is equal to F1<em>D1. The work the person must perform in the second case (with the ramp) is F2</em>D2. Since these works are equal, F1<em>D1=F2</em>D2. Solving for F2, the force used in the ramp scenario, F2=(F1*D1)/D2. Plug in 1m for D1 and 2m for D2 and you see that the force required to move the box up the ramp is half the force needed to lift the box 1m straight up.</p>
<p>Work equals F<em>d=4.2J. We must find the force needed to push the box up the 30 degree slope in order to solve for d, the distance. If we assume that the ramp is frictionless, we can find the force this way: the component of gravity parallel to the ramp is mg</em>sin(30) and this same force must be applied by the person in order to move the ramp upwards. Going back to W=F<em>d, d=W/F=W/(mg</em>sin(30))=4.2J/(0.6kg<em>9.8N/kg</em>sin(30))=1.43m.</p>
<p>Because the ball is spherical we can treat it like a convex mirror. Because convex mirrors are diverging mirrors, any image seen must be virtual, upright and reduced.</p>
<p>hm, for the first one, you’re right about W1=W2 but your equations are off. It would be W1=F1d1 and W2 = F2D2cos(x). Since it’s on a ramp, you have to take into consideration the angle of the ramp. You can find the angle of the ramp because it said work was the same, meaning height is the same. So the ramp is 2m long (hypotenuse) with a 1m shortest side (height). That means the angle must be 30deg. That means cos(x)=cos(30)=sqrt(3)/2. Plug that back into the equation for W2… you get W2 = F2D2Cos(30) = F1d1. d1 = 1 so F1=F2D2Cos(30). F2D2Cos30 = F2(2)(sqrt3)/2 so F1= F2sqrt(3). F2= F1/sqrt(3).</p>
<p>hope that was understandable…</p>
<p>physics,
thank you but my teacher said the answer was the ramp is half the force needed to lift the box 1m straight up. I since W=Fd is dealing with displacement, i don’t think you have to times cos30</p>
<p>the first one is 1/2</p>
<p>The worker is doing work against gravity, so lifting vertically = mgd
lifting up the ramp is mg(sin(X))d (vertical component of g down ramp)</p>
<p>where sinX = (opp/hyp) = (1/2)</p>
<p>mgd (up) : (1/2)mgd (ramp)…sooooo answer= 1/2 the force required to lift up</p>
<h1>2&3 i’m sure Dreadsilver got</h1>
<p>As long as the force is in the same direction as the displacement you can use W=F<em>d. Since force is in the same direction as the displacement up the ramp, you use the simple F</em>d. However, you *can *break the force into components.</p>
<p>When you break the force up into vertical and horizontal components, you must use two separate equations and add them to find the total work:
-----Horizontal: W=Fcos(theta)<em>horizontal displacement.
-----Vertical: W=Fsin(theta)</em>vertical displacement
Knowing that the triangle is the classic 1,2,root(3) type:
-----Horizontal: W=F(root(3)/2)<em>root(3)
-----Vertical: W=F(1/2)</em>1
Summing the two: Work = F(root(3)/2)<em>root(3) + F(1/2)</em>1=F(3/2) + F(1/2) = F*2.</p>
<p>Now, recall that this “F” is the force needed to move the box up the ramp (I call it F2). You set F2<em>2=F1</em>D1 and solve for F2: F2=F1<em>D1/(2) = F1</em>1/2</p>
<p>Of course, that 2 is the same two as the distance in the simpler method I used in my first post.</p>